Parallelogram is a quadrilateral whose opposite sides are parallel and opposite angles are equal. It is said to be special when they are either equiangular as in Rectangles, equilateral as in Rhombus or both as in Square.

## Properties of Special Parallelogram

Rectangle, rhombus and square are the special type of parallelograms.

Their properties are as follows:
Properties of Rectangle
i) Opposite sides are parallel to each other
ii) Opposite sides are congruent
iii) Opposite angles are equal
iv) Consecutive angles add up to 180 degrees, hence supplementary
v) Diagonals bisect each other
vi) The diagonals are equal in length
vii) All four angles at the vertices are right angle

Properties of Rhombus
i) Opposite sides are parallel
ii) Opposite sides are congruent
iii) Opposite angles are equal
iv) Consecutive angles add up to 180, hence supplementary
v) All four sides are equal
vi) The diagonals bisect each other cutting each other at equal halves
vii) Diagonals are perpendicular to each other making four right angles at the point of intersection
viii) Diagonals bisect each pair of opposite angles

Properties of Square
i) Opposite sides are parallel
ii) Opposite sides are congruent
iii) Opposite angles are equal
iv) Consecutive angles add up to 180, hence supplementary
v) Diagonals bisect each other
vi) Diagonals are perpendicular to each other
vii) All four sides are equal
viii) The diagonals are equal in length
ix) All four angles at the corner points are equal to 90 degrees
x) Diagonals bisect each pair of opposite angles

## Parallelogram Rectangle

A rectangle is a quadrilateral having corner angles equal to 90 degree each. Their pair of diagonals is equal.

Area of rectangle is the amount of space occupied by the four sides of the figure. It is calculated by the formula

Area = Length $\times$ Width

Perimeter of rectangle is the path covered by the rectangle. It is calculated by the formula

Perimeter = 2 $\times$ (Length + Width)

Diagonal of rectangle is the line joining the two opposite vertices of the rectangle. It is calculated by the formula

Diagonal = $\sqrt{Length^{2} + Width^{2}}$

Rectangle Example:

The sides of a rectangle are 8 inches and 4 inches respectively. Find out its area, perimeter and diagonal

Solution:

Area = Length $\times$ Width
Area = $8\ \times\ 4$
= $32$ square inches

Perimeter = $2\ \times$ (Length+Width)
= $2\ \times\ (8+4)$
= $2\ \times\ 12$

Perimeter = $24$ inches

Diagonal = $\sqrt{Length^{2} + Width^{2}}$
= $\sqrt{82 + 42}$
= $\sqrt{64 + 16}$
= $\sqrt{80}$ inches
Proof of Properties of Square:

Diagonals of rectangle are congruent

Proof:

Given: $PQRS$ is a rectangle

$PR$ and $QS$ are diagonals

In triangle $PQS$,

$QS^{2}$ = $PQ^{2}$ + $SP^{2}$ (Pythagoreas Theorem)

In triangle $PQR$,

$PR^{2}$ = $PQ^{2}$ + $QR^{2}$ (Pythagoreas Theorem)

$PS$ = $QR$ (Opposite sides of rectangle congruent)

Therefore, $QS^{2}$ = $PR^{2}$

$QS$ = $PR$

Hence, the diagonals of a rectangle are congruent

## Parallelogram Rhombus

Rhombus is a quadrilateral having all four sides equal. The pair of diagonals are perpendicular to each other and bisects each pair of opposite angles.

Area of rhombus is the amount of space occupied by the four sides of the figure. It is calculated by the formula

Area = $\frac{1}{2}$ $\times$ Product of the diagonals

Area = Length of Side $\times$ Perpendicular Distance between Parallel Lines

Area = $Side^{2}\ \times\ Sin(Supplementary\ Angle)$

Perimeter of rhombus is the path covered by the rhombus. It is calculated by the formula

Perimeter = Sum of the four sides

Rhombus Examples:

i) The diagonals of rhombus are 10 inches and 16 inches respectively. Find out its area and perimeter

Solution:

Area = $\frac{1}{2}$ $\times$ Product of the diagonals

Area = $\frac{1}{2}$ $\times\ 10\ \times\ 16$

= $80$ square inches

Perimeter = $4\ \times$ $\sqrt{(\frac{10}{2})^{2}}$ + $(\frac{16}{2})^{2}$

= $4\ \times\ \sqrt{(25 + 64)}$

= $4\ \times\ \sqrt{89}$inches

ii) The perimeter of a rhombus is $160$ feet and one of its diagonal has a length of $20$ feet. Find the area of the rhombus

Solution:

Length of each side of Rhombus = $\frac{160}{4}$ feet = $40$ feet

$402$ = $(\frac{20}{2})^{2}$ + $(\frac{length\ of\ other\ diagonal}{2})^{2}$

Length of the other diagonal = $2\ \times\ \sqrt{(1600-100)}$ = $77.46$ feet

Area = $\frac{1}{2}$ $\times$ Product of the diagonals

= $\frac{1}{2}$ $\times\ 20\ \times\ 77.46$ = $774.6$ square feet
Proof of properties of Rhombus:

i) Diagonals of a Rhombus are perpendicular to each other

Proof: Let us consider triangles $POQ$ and $QOR$. We know that the diagonals bisect each other, so $PO$ = $OR$

Given: $PQRS$ is a Rhombus

$PO$ = $OR$ (Diagonals bisect each other)

$QO$ = $QO$ (reflexive property of congruency)

$PQ$ = $QR$ (Sides of Rhombus are equal as per properties of Rhombus)

Therefore, Triangle $POQ$ Congruent Triangle $QOR$ (By $SSS$ Property)

It is known to us that corresponding angles in congruent triangles are congruent. We see that angles $POQ$ and $QOR$ are corresponding angles. Thus, angle $POQ$ congruent angle $QOR$.

$O$ lies on the straight line $PR$.

Hence,

Angle $POR$ + angle $QOR$ = $180$degrees (supplementary angles)

Angle $POR$ + Angle $POR$ = $180$degrees

$2\ \times$ Angle $POR$ = $180$ (Addition property)

Angle $POR$ = $90$ (Division property)

Angle $QOR$ = Angle $POR$ = $90$ degrees (As these two angles are congruent)

Therefore, the diagonals of a Rhombus are perpendicular to each other.
ii) Diagonals of a Rhombus bisect each other

Proof: Let us consider triangles $POQ$ and $ROS$

Angle $QPO$ = Angle $SRO$ (Alternate Interior Angles as $PQ$ parallel to $SR$)

Angle $PQO$ = Angle $RSO$ (Alternate Interior Angles as $PQ$ parallel to $SR$)

Angle $POQ$ = Angle $SOR$ (Vertically opposite angles)

Therefore triangle $POQ$ Congruent triangle $SOR$ ($ASA$ property of congruency)

Thus, $PO$ = $OR$ (By CPCT)
$QO$ = $OS$ (By CPCT)

Hence, diagonals of a Rhombus bisect each other

## Parallelogram Square

Square is a quadrilateral having all four sides are equal, diagonals are equal and perpendicular bisectors of each other and the corner angles are of $90$ degree each.

Area of square is the region bounded by its four equal sides. It is calculated by the following formula

Area = Side$^{2}$

Area = $\frac{1}{2}$ $\times$ Diagonal$^{2}$

Perimeter is the path covered by the square. It is calculated by the following formula

Perimeter = $4\ \times$ Length of each side

Diagonal = $\sqrt{2}\ \times$ Side
Square Problems:

The length of side of a square is $4cm$. Find its area, perimeter and diagonal

Solution:

Area = Side $\times$ Side
= $4\ \times\ 4$
= $16$ square cms
Perimeter = $4\ \times$ Side
= $4\ \times\ 4$
= $16$cms

Diagonal = $\sqrt{2}\ \times$ Side
= $4 \sqrt{2}$cms
Proof of Properties of Square

If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a square

Given: $PQRS$ is a parallelogram

$PR$ = $QS$ and $PR$ perpendicular $QS$

$PO$ = $PO$ (Reflexive)

Angle $POQ$ = Angle $POS$ (Each $90$ degree)

$OQ$ = $OS$(Parallelogram property)

Triangle $POQ$ congruent Triangle $POS$ (SAS postulate)

$PQ$ = $PS$ (CPCT)

$PQ$ = $RS$ (Parallelogram properties)

$PR$ = $QS$(Parallelogram properties)

Hence, $PQ$ = $QR$ = $RS$ = $PS$

$PQ$ = $PQ$(Reflexive)

$PS$ = $QR$ (Parallelogram property)

$PR$ = $QS$ (Given)

Triangle $PQS$ congruent Triangle $QPR$ (SSS Postulate)

Angle $SPQ$ = Angle $RQP$ (CPCT)

Angle $SPQ$ + Angle $RQP$ = 180 (Interior angles on the same side of the transversal)

$2\ \times$ Angle $SPQ$ = $180$ (Addition Property)

Angle $SPQ$ = Angle $RQP$ = $90$ (Division property)